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let-A-n-cos-2-n-x-x-2-3-2-dx-1-calculate-A-n-interms-of-n-2-find-nsture-of-the-serie-A-n-and-n-n-A-n-

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Question Number 65679 by mathmax by abdo last updated on 01/Aug/19
let A_n =∫_(−∞) ^(+∞) ((cos(2^n x))/((x^2 +3)^2 ))dx 1) calculate A_n interms of n 2)find nsture of the serie ΣA_n and Σn^n A_n
$${let}\:\:{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}^{{n}} {x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right){find}\:{nsture}\:{of}\:{the}\:{serie}\:\Sigma{A}_{{n}} \:\:\:\:{and}\:\Sigma{n}^{{n}} \:{A}_{{n}} \\ $$
Commented by mathmax by abdo last updated on 02/Aug/19
2) Σ_(n=0) ^∞ A_n =(π/6)Σ_(n=0) ^∞ 2^n e^(−(√3)2^n ) +((π(√3))/(18)) Σ_(n=0) ^∞ e^(−(√3)2^n ) those series converges ⇒Σ A_n converges we can verify that lim_(n→+∞) n^n A_n ≠0 ⇒Σ A_n diverges
$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \:=\frac{\pi}{\mathrm{6}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\mathrm{2}^{{n}} \:{e}^{−\sqrt{\mathrm{3}}\mathrm{2}^{{n}} } \:\:+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−\sqrt{\mathrm{3}}\mathrm{2}^{{n}} } \\ $$$${those}\:{series}\:\:{converges}\:\:\Rightarrow\Sigma\:{A}_{{n}} \:{converges} \\ $$$${we}\:{can}\:{verify}\:{that}\:{lim}_{{n}\rightarrow+\infty} {n}^{{n}} \:{A}_{{n}} \:\neq\mathrm{0}\:\Rightarrow\Sigma\:{A}_{{n}} {diverges} \\ $$
Commented by mathmax by abdo last updated on 02/Aug/19
1) we have A_n =∫_(−∞) ^(+∞) ((cos(2^n x))/((x^2 +3)^2 ))dx ⇒A_n =Re(∫_(−∞) ^(+∞) (e^(i2^n x) /((x^2 +3)^2 ))dx) let ϕ(z) =(e^(i2^n z) /((z^2 +3)^2 )) poles of ϕ? ϕ(z) =(e^(i2^n z) /((z−i(√3))^2 (z+i(√3))^2 )) the poles of ϕ are +^− i(√3)(doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,i(√3)) Res(ϕ,i(√3)) =lim_(z→i(√3)) (1/((2−1)!)){(z−i(√3))^2 ϕ(z)}^((1)) =lim_(z→i(√3)) {(e^(i2^n z) /((z+i(√3))^2 ))}^((1)) =lim_(z→i(√3)) ((i2^n e^(i2^n z) (z+i(√3))^2 −2(z+i(√3))e^(i2^n z) )/((z+i(√3))^4 )) =lim_(z→i(√3)) ((i2^n e^(i2^n z) (z+i(√3))−2e^(i2^n z) )/((z+i(√3))^3 )) =((i2^n e^(i2^n (i(√3))) (2i(√3))−2e^(i2^n (i(√3))) )/((2i(√3))^3 )) =((−2(√3)2^n e^(−2^n (√3)) −2 e^(−2^n (√3)) )/((−8i)(3(√3)))) =(((2(√3)2^n +2)e^(−2^n (√3)) )/(24i(√3))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((((√3)2^n +1)e^(−2^n (√3)) )/(12i(√3))) =(π/(6(√3)))( (√3)2^n +1)e^(−2^n (√3)) ⇒ A_n =(π/(18)){3. 2^n +(√3))e^(−(√3)2^n )
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{2}^{{n}} {x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\Rightarrow{A}_{{n}} ={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\mathrm{2}^{{n}} {x}} }{\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{{e}^{{i}\mathrm{2}^{{n}} {z}} }{\left({z}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:\varphi? \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{i}\mathrm{2}^{{n}} {z}} }{\left({z}−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i}\sqrt{\mathrm{3}}\left({doubles}\right) \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\:\:\left\{\frac{{e}^{{i}\mathrm{2}^{{n}} {z}} }{\left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\:\frac{{i}\mathrm{2}^{{n}} {e}^{{i}\mathrm{2}^{{n}} {z}} \left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\sqrt{\mathrm{3}}\right){e}^{{i}\mathrm{2}^{{n}} {z}} }{\left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\:\frac{{i}\mathrm{2}^{{n}} {e}^{{i}\mathrm{2}^{{n}} {z}} \left({z}+{i}\sqrt{\mathrm{3}}\right)−\mathrm{2}{e}^{{i}\mathrm{2}^{{n}} {z}} }{\left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$$=\frac{{i}\mathrm{2}^{{n}} \:{e}^{{i}\mathrm{2}^{{n}} \left({i}\sqrt{\mathrm{3}}\right)} \left(\mathrm{2}{i}\sqrt{\mathrm{3}}\right)−\mathrm{2}{e}^{{i}\mathrm{2}^{{n}} \left({i}\sqrt{\mathrm{3}}\right)} }{\left(\mathrm{2}{i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\:\:=\frac{−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{2}^{{n}} \:{e}^{−\mathrm{2}^{{n}} \sqrt{\mathrm{3}}} −\mathrm{2}\:{e}^{−\mathrm{2}^{{n}} \sqrt{\mathrm{3}}} }{\left(−\mathrm{8}{i}\right)\left(\mathrm{3}\sqrt{\mathrm{3}}\right)} \\ $$$$=\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}\mathrm{2}^{{n}} \:+\mathrm{2}\right){e}^{−\mathrm{2}^{{n}} \sqrt{\mathrm{3}}} }{\mathrm{24}{i}\sqrt{\mathrm{3}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\left(\sqrt{\mathrm{3}}\mathrm{2}^{{n}} \:+\mathrm{1}\right){e}^{−\mathrm{2}^{{n}} \sqrt{\mathrm{3}}} }{\mathrm{12}{i}\sqrt{\mathrm{3}}} \\ $$$$=\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}}\left(\:\:\sqrt{\mathrm{3}}\mathrm{2}^{{n}} \:+\mathrm{1}\right){e}^{−\mathrm{2}^{{n}} \sqrt{\mathrm{3}}} \:\:\:\Rightarrow\:{A}_{{n}} =\frac{\pi}{\mathrm{18}}\left\{\mathrm{3}.\:\mathrm{2}^{{n}} \:+\sqrt{\mathrm{3}}\right){e}^{−\sqrt{\mathrm{3}}\mathrm{2}^{{n}} } \\ $$

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